Question

A block of mass m is at rest on a rough inclined plane of angle of inclination θ . If coefficient of friction between the block and the inclined plane is required to move the block on the plane is

how the answer is   mg[sinθ - μ cosθ]  .

 

Joshi sir comment

i think one line is missing in the question.

Read 2 Solution.

I think something u have skipped in the question

 if u r asking for the force F required to move the block then d first step is to resolve mg in its two component & then make balance condition R=mgcosθ    & F= mgsinθ + f            f=uR

 F= mg sinθ + µR =mg sinθ+ µmg cosθ =mg( sinθ+cosθ)  as d body is moving up therefore friction f will act opposite to applied force in the direction of mgsinθ thus sign in b/w should be + , whereas u have used -ve 

SARIKA SHARMA 11 year ago is this solution helpfull: 2 0

 if u asking to move the body  up then the solution submitted will be correct, But if u are asking for just to make it move or to move downward then the sign vyl be changed then balance conditn will be

 F = mgsinθ-f = mg (sin θ -ucosθ)     here f=uR & r=mgcosθ   as force to move it is along sin component thus sign will be same 7 as friction f is acting opposite thus it has -ve sign . thus both ans are applicable depending upon the force is req in which direction

Best SolutionSARIKA SHARMA 11 year ago is this solution helpfull: 2 0

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